question_answer

The 'spin-only' magnetic moment [in units of Bohr magneton,\[({{\mu }_{\beta }})]\]of\[N{{i}^{2+}}\]in aqueous solution would be (Atomic number of\[Ni=28\])     AIEEE  Solved  Paper-2006

A) 2.84       

B)        4.90       

C)        0                             

D)        1.73

Correct Answer: A

Solution :

\[N{{i}^{2+}}=[Ar]3{{d}^{8}}\] Number of unpaired electrons = 2 Hence, magnetic moment\[=\sqrt{n(n+2)}=\sqrt{8}=2.84\]